Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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If deg⁡xiP<∣Si∣ for each i and P vanishes on S1×⋯×Sn, then P=0

Statement

Let F be a field, let P∈F[x1,…,xn], and let finite sets S1,…,Sn⊆F satisfy

deg⁡xiP<∣Si∣(1≤i≤n).

If P vanishes at every point of S1×⋯×Sn, then P=0.

Facts & Assumptions

Given: a field F, finite subsets S1,…,Sn⊆F, and a polynomial P∈F[x1,…,xn] with deg⁡xiP<∣Si∣ for every i, vanishing on S1×⋯×Sn.

[F2]

Every polynomial has a finite monomial expansion, so in particular one may write P=∑j=0dPj(x1,…,xn−1)xnj with d=deg⁡xnP (Monomials, coefficients, degree in each variable and total degree in F[x1,…,xn]).

Proof

technique · induction
1.1F1

[base] If n=1, then P is a univariate polynomial of degree below ∣S1∣ vanishing at every point of S1. By [F1] a nonzero polynomial of that degree cannot have so many roots, so P=0.

1.2given

[ih] Assume the statement is known for polynomials in n−1 variables.

1.3F2

Write P=∑j=0dPj(x1,…,xn−1)xnj as in [F2], where d=deg⁡xnP<∣Sn∣.

2.1F1step 1.3

Fix (a1,…,an−1)∈S1×⋯×Sn−1. Then the univariate polynomial Q(y):=P(a1,…,an−1,y) has degree at most d<∣Sn∣ and vanishes on all of Sn, so [F1] gives Q=0. Therefore every coefficient Pj(a1,…,an−1) is 0.

3.1step 1.2step 2.1discharge-induction∎

Since the point (a1,…,an−1) was arbitrary, every coefficient polynomial Pj vanishes on S1×⋯×Sn−1. Its degree in each variable is still below the corresponding ∣Si∣, so the induction hypothesis gives Pj=0 for every j. Hence P=0.

Remarks

  • The inequalities are strict. The companion page's counterexample is exactly the boundary case where equality holds in one variable.

Depends on

Used by

Dependency tree · two levels

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Sources